Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.
1 Page
IRWIN 9e 14_FE-2
Course: EE 2120, Fall 2008School: LSU
Rating:
Document Preview
Basic Irwin, Engineering Circuit Analysis, 9/E 1 SOLUTION: The correct answer is c. 4 I ( s) 2 sI ( s ) 2 I ( s ) s 4 V s ( s ) 2 s 2 I ( s ) s V0 ( s ) 2 I ( s ) V (s) (s) I 0 2 4 V ( s) Vs ( s ) 2 s 2 0 s 2 V0 ( s ) s 2 Vs ( s ) s s 2 Vs ( s ) Chapter 14: Application of the Laplace Transform To Circuit Analysis Problem 14.FE-2
Unformatted Document Excerpt
Coursehero >> Louisiana >> LSU >> EE 2120Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
Basic Irwin, Engineering Circuit Analysis, 9/E
1
SOLUTION: The correct answer is c.
4 I ( s) 2 sI ( s ) 2 I ( s ) s 4 V s ( s ) 2 s 2 I ( s ) s V0 ( s ) 2 I ( s ) V (s) (s) I 0 2 4 V ( s) Vs ( s ) 2 s 2 0 s 2 V0 ( s ) s 2 Vs ( s ) s s 2 Vs ( s )
Chapter 14: Application of the Laplace Transform To Circuit Analysis
Problem 14.FE-2
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more.
Course Hero has millions of course specific materials providing students with the best way to expand
their education.
Below is a small sample set of documents:
Below is a small sample set of documents:
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is a. s4 I s (s) I 0 ( s) s4 3 s I 0 (s) s( s 4) 2 I s (s) s 4s 3Chapter 14: Application of the Laplace Transform To Circuit AnalysisProblem 14.FE-3
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is b. KVL:4s 4 s 2 I (s) s s 1 2I (s) 4s 2 ( s 2 1)( s 2 2 s 4)Chapter 14: Application of the Laplace Transform To Circuit AnalysisProblem 14.FE-4
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is d. KVL: Vs ( s ) 4 I ( s) 2 sI ( s )Vs ( s ) (2 s 4) I ( s ) V0 ( s) 2 sI ( s) V (s) I (s) 0 2sV s ( s ) ( 2 s 4) V0 ( s ) 2sV0 ( s ) s Vs ( s ) s 2Chapter 14: Applicatio
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
(s)(s)-(s) =(s) =(s) =(s)(s)(s)(s)o (s)(s) =(s)(s)-(s)(s)(s)(s)(s)(s)(s)(s)(s)(s)-
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is b. By observation, the waveform has odd symmetry. Therefore, an = 0 for all n due to odd symmetry.Chapter 15: Fourier Analysis TechniquesProblem 15.FE-1
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is d. By observation, the waveform has half-wave symmetry. Therefore, bn =0 for n even due to half-wave symmetry and bn is nonzero for n odd.Chapter 15: Fourier Analysis Techni
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is a. j V0 (n ) 1 j Vs (n ) V0 (0) 0 j 2 30 V0 ( ) 1 j 2 0 V0 ( ) 8.5426.57V j 4 30 V0 (2 ) 1 j 4 2 0 V0 (2 ) 4.6314.04V j 6 30 V0 (3 ) 1 j 6 3 0 V0 (3 ) 3.149.46V j8 30 V0 (
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is b.I (n ) I DC Vs 2 jn3.7720 10 A 2 1060 I1 2.34 2.05 A 2 j 3.77 445 I3 0.35 34.97 A 2 j11.31 i (t ) 10 2.34 cos(377t 2.05) 0.35 cos(1131t 34.97) A VI P VDC I DC n n cos(
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is b.a0 a0 1 1 10 dt 4 2 dt 40 2241 2 10t 0 1 2t 4 2 4 4 1 1 a 0 (20) 2(4) (2)(2) 4 4 a 0 4VChapter 15: Fourier Analysis TechniquesProblem 15.FE-5
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120